Question

Six point masses each of mass $m$ are placed at the vertices of a regular hexagon of side $l$. The force acting on any of the masses is

• A. $\frac{G{m}^2}{l^2}\left[\frac{5}{4}+\frac{1}{\sqrt{3}}\right]$
• B. $\frac{G{m}^2}{l^2}\left[\frac{3}{4}+\frac{1}{\sqrt{3}}\right]$
• C. $\frac{G{m}^2}{l^2}\left[\frac{5}{4}-\frac{1}{\sqrt{3}}\right]$
• D. $\frac{G{m}^2}{l^2}\left[\frac{3}{4}-\frac{1}{\sqrt{3}}\right]$

Answer 1

Answer: (A)

from figure,
$AC=AM+MC=2AM=2lcos30^{\circ }=2l\frac{\sqrt{3}}{2}=\sqrt{3}l$
Similarly, $AE=\sqrt{3}l$,
$AD=AO+ON+ND=lsin30^{\circ }+l+lsin30^{\circ }$
$=l\times \frac{1}{2}+l+\times \frac{1}{2}=2l$
$AB=AF=l$
Force on mass $m$ at $A$ due to mass $m$ at $B$ is
$F_{AB}=\frac{Gmm}{{\left(AB\right)}^2}=\frac{Gmm}{l^2}$ along $AB$
Force on mass $m$ at $A$ due to mass $m$ at $C$ is
$F_{AC}=\frac{Gmm}{{\left(AC\right)}^2}=\frac{Gmm}{l^2}$ along $AC$
Force on mass $m$ at $A$ due to mass $m$ at $D$ is
$F_{AD}=\frac{Gmm}{{\left(AD\right)}^2}=\frac{Gmm}{l^2}$ along $AD$
Force on mass $m$ at $A$ due to mass $m$ at $E$ is
$F_{AE}=\frac{Gmm}{{\left(AE\right)}^2}=\frac{Gmm}{l^2}$ along $AE$
Force on mass $m$ at $A$ due to mass $m$ at $F$ is
$F_{AF}=\frac{Gmm}{{\left(AF\right)}^2}=\frac{Gmm}{l^2}$ along $AF$
Resultant force due to $F_{AB}$ and $F_{AF}$ is
$F_{R_1}=\sqrt{F_{AB}^2+F_{AF}^2+2F_{AB}{F}_{AF}cos120^{\circ }}$
$=\sqrt{{\left(\frac{G{m}^2}{l^2}\right)}^2+{\left(\frac{G{m}^2}{l^2}\right)}^2+2{\left(\frac{G{m}^2}{l^2}\right)}^2{\left(\frac{G{m}^2}{l^2}\right)}^2\left(-\frac{1}{2}\right)}$
$=\frac{G{m}^2}{l^2}$ along $AD$
Resultant force due to $F_{AC}$ and $F_{AE}$ is
$F_{R_2}=\sqrt{F_{AC}^2+F_{AE}^2+2F_{AC}{F}_{AE}cos60^{\circ }}$
$=\sqrt{{\left(\frac{G{m}^2}{3l^2}\right)}^2+{\left(\frac{G{m}^2}{3l^2}\right)}^2+2{\left(\frac{G{m}^2}{3l^2}\right)}^2{\left(\frac{G{m}^2}{3l^2}\right)}^2\left(\frac{1}{2}\right)}$
$=\frac{\sqrt{3}G{m}^2}{3l^2}=\frac{G{m}^2}{\sqrt{3}{l}^2}$ along $AD$
Net force on mass $m$ along $AD$ is
$F_R=F_{R_1}+F_{R_2}+F_{AD}=\frac{G{m}^2}{l^2}+\frac{G{m}^2}{\sqrt{3}{l}^2}+\frac{G{m}^2}{4l^2}$
$=\frac{G{m}^2}{l^2}\left[1+\frac{3}{\sqrt{3}}+\frac{1}{4}\right]$
$=\frac{G{m}^2}{l^2}\left[\frac{5}{4}+\frac{1}{\sqrt{3}}\right]$